- Consider a circular ring of radius 'a' which carries a uniform line charge density ρL as shown in figure.
- We need to find out electric field at a point P (0, 0, h) on the z axis (z > 0).
- Electric field intensity (E) due to any line charge (ρL) in general is given as:
In this case,
dl = a dφ (Since the differential part dl is a differential arc)
R2 = a2 + h2
Consider the triangle shown in the above figure
a aρ + R = h az → R = - a aρ + h az
aR = R / | R |
aR = - a aρ + h az / ( a2 + h2 )1/2
Substituting all these values in the above equations, the electric field intensity E becomes:
- For every element dl there is a corresponding element diametrically opposite that gives an equal but opposite dEρ so that the two contributions cancel each other.
Hence contribution along aρ due to symmetry adds up to zero.
- Therefore the final electric field intensity at point (0, 0, h) has only z component.
- Hence Electric field intensity (E) due to a circular ring (of radius a carrying a uniform charge ρL) placed on the x-y plane and if the point of interest is any point on z axis, then it is given as:
- Similarly if the ring is placed on x-z plane and point of interest is any point on y- axis, then E is given as:
- Introduction To Electrostatics.
- Coulomb's law.
- Electric Field Intensity (E).
- Electric Lines Of Forces /Streamlines / Electric Flux (ψ) .
- Electric Flux Density (D).
- Electric Field Intensity Due To a Finite and Infinite Line Charge.
- Electric Field Intensity Due To a Infinite Sheet Charge.
- Electric Field Intensity Due To a Circular Ring Charge.
- Electric Field Intensity Due To a Circular Disk Charge.
- Numericals / Solved Examples - Electric Force and Field Intensity.
- Numericals / Solved Examples - Electric Field Intensity - Line, Surface and Mixed Charge Configuration.
Your suggestions and comments are welcome in this section. If you want to share something or if you have some stuff of your own, please do post them in the comments section.